Flux Across Surfaces - Multivariable Calculus

1. Consider a vector field $\mathbf{F}(x,y,z) = (P(x,y,z), Q(x,y,z), R(x,y,z))$ where $P$, $Q$, and $R$ are continuously differentiable functions. If the flux of $\mathbf{F}$ through every closed surface in a region is zero, and you need to compute the flux through an open surface $S$ with boundary curve $C$, which approach is most appropriate?
   Answer: Since $\nabla \cdot \mathbf{F} = 0$ in the region, the flux through $S$ equals the flux through any other surface with the same boundary $C$ When you encounter a vector field where the flux through every closed surface is zero, you're dealing with a solenoidal (divergence-free) field. This condition tells you that $\nabla \cdot \mathbf{F} = 0$ throughout the region, which has profound implications for flux calculations. The key insight is that when $\nabla \cdot \mathbf{F} = 0$, the flux through any surface depends only on its boundary, not the specific surface itself. If two surfaces $S_1$ and $S_2$ share the same boundary curve $C$, then by the divergence theorem applied to the closed surface formed by $S_1$ and $-S_2$, we get $\iint_{S_1} \mathbf{F} \cdot d\mathbf{S} - \iint_{S_2} \mathbf{F} \cdot d\mathbf{S} = \iiint_V \nabla \cdot \mathbf{F} \, dV = 0$. Therefore, both surfaces have equal flux. Answer D captures this principle perfectly. A is incorrect because Stokes' theorem relates circulation (line integrals) to curl, not flux to divergence. B misses the simplification opportunity entirely—direct parametrization would be unnecessarily complex when we can exploit the divergence-free property. C has the right idea about using closed surfaces but describes an overly complicated approach when the simpler surface-substitution method works directly. Study tip: When you see "flux through every closed surface is zero," immediately think "divergence-free field" and "surface independence." This means you can always replace a complicated surface with a simpler one having the same boundary—often a flat disk or plane region.
2. Consider the hemisphere $S: x^2 + y^2 + z^2 = 9$ with $z \geq 0$, oriented with outward normal vectors. The flux of the vector field $\mathbf{F}(x,y,z) = (x^3, y^3, z^3)$ through this hemisphere (not including the circular base) can be computed most efficiently using which approach?
   Answer: Applying divergence theorem to the entire hemisphere volume, then subtracting the flux through the circular base Since $\nabla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 = 3(x^2 + y^2 + z^2)$, which is constant ($= 27$) on the sphere, the divergence theorem is most efficient. We compute the flux through the closed hemisphere (including the base), then subtract the flux through the base. The total flux through the closed surface is $\iiint_V 27 \, dV = 27 \cdot \frac{2\pi}{3} \cdot 3^3 = 486\pi$. The flux through the base is $\iint_{x^2+y^2 \leq 9} z^3 \, dx \, dy = 0$ (since $z = 0$ on the base). Choice A would work but is computationally intensive. Choice C is not a standard approach for surface integrals. Choice D incorrectly applies Green's theorem, which is for line integrals.
3. A surface $S$ is defined parametrically by $\mathbf{r}(u,v) = (u\cos v, u\sin v, u^2)$ for $0 \leq u \leq 2$ and $0 \leq v \leq \pi$. If the vector field $\mathbf{F}(x,y,z) = (0, 0, xyz)$ flows through this surface, and the surface is oriented using the normal vector $\mathbf{r}_u \times \mathbf{r}_v$, what is the flux of $\mathbf{F}$ through $S$?
   Answer: $\int_0^\pi \int_0^2 u^5 \cos v \sin v \, du \, dv$ First, we find $\mathbf{r}_u = (\cos v, \sin v, 2u)$ and $\mathbf{r}_v = (-u\sin v, u\cos v, 0)$. Then $\mathbf{r}_u \times \mathbf{r}_v = (-2u^2\cos v, -2u^2\sin v, u)$. On the surface, $\mathbf{F} = (0, 0, u\cos v \cdot u\sin v \cdot u^2) = (0, 0, u^4\cos v \sin v)$. The flux integral is $\iint_S \mathbf{F} \cdot d\mathbf{S} = \int_0^\pi \int_0^2 (0, 0, u^4\cos v \sin v) \cdot (-2u^2\cos v, -2u^2\sin v, u) \, du \, dv = \int_0^\pi \int_0^2 u^5\cos v \sin v \, du \, dv$. Choice A forgets one factor of $u$. Choice C is missing $u^2$. Choice D has an extra factor of $u$.
4. The flux of $\mathbf{F}(x,y,z) = (y^2, xz, z^2-x)$ through the triangular surface with vertices at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$, oriented so that the normal vector points away from the origin, is being computed. Which statement about this computation is correct?
   Answer: The surface can be parametrized as $\mathbf{r}(u,v) = (1-u-v, u, v)$ with $u,v \geq 0$ and $u+v \leq 1$, and the normal vector $\mathbf{r}_u \times \mathbf{r}_v = (1,1,1)$ points away from the origin When computing flux through a surface, you need three key components: a parametrization of the surface, the correct normal vector with proper orientation, and the flux integral setup. The triangular surface has vertices at $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$. These points all satisfy $x + y + z = 1$, confirming the surface lies on this plane. The parametrization $\mathbf{r}(u,v) = (1-u-v, u, v)$ correctly represents this triangle: when $u = v = 0$, you get $(1,0,0)$; when $u = 1, v = 0$, you get $(0,1,0)$; and when $u = 0, v = 1$, you get $(0,0,1)$. To find the normal vector, compute $\mathbf{r}_u = (-1,1,0)$ and $\mathbf{r}_v = (-1,0,1)$, giving $\mathbf{r}_u \times \mathbf{r}_v = (1,1,1)$. To check orientation, pick a point on the surface like $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$ and verify that this point plus the normal vector $(1,1,1)$ moves away from the origin. Indeed, $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}) + (1,1,1) = (\frac{4}{3}, \frac{4}{3}, \frac{4}{3})$ is farther from the origin. Choice A incorrectly claims $(1,1,1)$ points toward the origin. Choice B shows an incorrect flux integral computation. Choice C incorrectly suggests using the unit normal $\frac{(1,1,1)}{\sqrt{3}}$ instead of $(1,1,1)$ itself. Choice D correctly identifies both the parametrization and proper orientation. Study tip: Always verify normal vector orientation by checking whether adding the normal to a surface point moves you toward or away from the reference point.
5. Consider the vector field $\mathbf{F}(x,y,z) = (2x, -y, z^2)$ and the surface $S$ consisting of the portion of the paraboloid $z = x^2 + y^2$ for $0 \leq z \leq 4$, oriented with upward-pointing normal vectors. If the flux of $\mathbf{F}$ across $S$ is computed using the surface integral $\iint_S \mathbf{F} \cdot d\mathbf{S}$, which expression correctly represents this flux?
   Answer: $\iint_D (-4x^2 + 2xy + (x^2 + y^2)^2) \, dx \, dy$ where $D: x^2 + y^2 \leq 4$ For the surface $z = g(x,y) = x^2 + y^2$, the upward normal vector is $\mathbf{n} = \frac{(-g_x, -g_y, 1)}{\sqrt{1 + g_x^2 + g_y^2}} = \frac{(-2x, -2y, 1)}{\sqrt{1 + 4x^2 + 4y^2}}$. The flux integral becomes $\iint_D \mathbf{F} \cdot \mathbf{n} \, dS = \iint_D (2x, -y, (x^2+y^2)^2) \cdot (-2x, -2y, 1) \, dx \, dy = \iint_D (-4x^2 + 2xy + (x^2+y^2)^2) \, dx \, dy$. Choice A has the wrong sign for the y-component. Choice B has both wrong signs and includes the unnecessary $\sqrt{1+4x^2+4y^2}$ factor. Choice D has the wrong sign for the xy term.
6. Consider the surface $S$ defined by $x^2 + y^2 + z^2 = 1$ with $z \geq 0$ (upper hemisphere), and let $\mathbf{F}(x,y,z) = (z, 0, x)$. To find the flux of $\mathbf{F}$ through $S$ oriented with outward normal vectors, a student computes $\nabla \cdot \mathbf{F} = 0$ and concludes the flux is zero by the divergence theorem. What is wrong with this reasoning?
   Answer: The divergence theorem applies to closed surfaces, but $S$ is not closed; the circular base at $z = 0$ must be included When you encounter flux problems involving the divergence theorem, always check whether the surface satisfies the theorem's requirements. The divergence theorem states that $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_E \nabla \cdot \mathbf{F} \, dV$, but it only applies to closed surfaces that completely enclose a region. The student's error lies in applying the divergence theorem to an incomplete surface. While the divergence calculation $\nabla \cdot \mathbf{F} = \frac{\partial z}{\partial x} + \frac{\partial 0}{\partial y} + \frac{\partial x}{\partial z} = 0$ is correct, the upper hemisphere alone doesn't form a closed surface—it has an open circular boundary at $z = 0$. To use the divergence theorem, you must include the disk $x^2 + y^2 \leq 1, z = 0$ to create a complete closed surface enclosing the solid hemisphere. Looking at the wrong answers: (B) is incorrect because $\mathbf{F} = (z, 0, x)$ is continuously differentiable everywhere, including on the hemisphere. (C) contains a calculation error—the divergence is indeed zero, not 1, since $\frac{\partial x}{\partial z} = 0$. (D) is wrong because outward normal orientation is perfectly valid and standard for flux calculations through hemispheres. Answer: (A) Study tip: Before applying the divergence theorem, always verify the surface is closed. If you have a partial surface like a hemisphere, cap, or portion of a cylinder, you'll need to add the missing pieces to create a closed boundary, or use direct surface integration instead.
7. A surface $S$ is given by the graph of $z = f(x,y)$ over a region $D$ in the xy-plane, where $f$ has continuous partial derivatives. The flux of $\mathbf{F}(x,y,z) = (P, Q, R)$ through $S$ with upward orientation is computed as $\iint_D (-P f_x - Q f_y + R) \, dx \, dy$. If instead we want the flux through the same surface but with downward orientation, the integral becomes:
   Answer: $\iint_D (P f_x + Q f_y - R) \, dx \, dy$ For a surface $z = f(x,y)$ with upward orientation, the unit normal vector is $\mathbf{n} = \frac{(-f_x, -f_y, 1)}{\sqrt{1 + f_x^2 + f_y^2}}$ and $d\mathbf{S} = \mathbf{n} \, dS = (-f_x, -f_y, 1) \, dx \, dy$. The flux integral becomes $\iint_D \mathbf{F} \cdot d\mathbf{S} = \iint_D (P, Q, R) \cdot (-f_x, -f_y, 1) \, dx \, dy = \iint_D (-P f_x - Q f_y + R) \, dx \, dy$. For downward orientation, we use $-\mathbf{n}$, so $d\mathbf{S} = (f_x, f_y, -1) \, dx \, dy$, giving $\iint_D (P, Q, R) \cdot (f_x, f_y, -1) \, dx \, dy = \iint_D (P f_x + Q f_y - R) \, dx \, dy$. Choice B incorrectly includes the magnitude factor $\sqrt{1 + f_x^2 + f_y^2}$. Choice C incorrectly divides by this factor. Choice D has the wrong sign on the $R$ term.
8. Two surfaces $S_1$ and $S_2$ have the same boundary curve $C$ but different orientations. If $S_1$ is oriented with normal vectors $\mathbf{n}_1$ and $S_2$ is oriented with normal vectors $\mathbf{n}_2 = -\mathbf{n}_1$, and if $\Phi_1$ and $\Phi_2$ represent the flux of a vector field $\mathbf{F}$ through $S_1$ and $S_2$ respectively, which relationship must hold?
   Answer: $\Phi_1 + \Phi_2 = 0$ if the surfaces $S_1$ and $S_2$ are geometrically identical, regardless of the vector field If $S_1$ and $S_2$ are geometrically identical surfaces with opposite orientations ($\mathbf{n}_2 = -\mathbf{n}_1$), then $\Phi_2 = \iint_{S_2} \mathbf{F} \cdot \mathbf{n}_2 \, dS = \iint_{S_1} \mathbf{F} \cdot (-\mathbf{n}_1) \, dS = -\iint_{S_1} \mathbf{F} \cdot \mathbf{n}_1 \, dS = -\Phi_1$. Therefore $\Phi_1 + \Phi_2 = 0$ regardless of the vector field, as long as the surfaces are geometrically identical. Choice A is incorrect because it doesn't require the surfaces to be identical. Choice B is wrong because identical surfaces with opposite orientations give opposite flux values, not equal ones. Choice D is incorrect because even with $\nabla \cdot \mathbf{F} = 0$, different surfaces can have different flux values; the condition only guarantees that flux depends on boundary, not that it's equal for opposite orientations.

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